Yt14-hdu-found the right x in 0-100.

Problem Descriptionnow,given the equation 8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 = = Y,can you find its solution between 0 and 100 ;
Now try your lucky. Inputthe first line of the input contains an integer T (1<=t<=100) which means the number of test cases. Then T lines follow with each line have a real number Y (Fabs (Y) <= 1e10) outputfor Each test case, you should just output O NE real number (accurate up to 4 decimal places), which are the solution of the Equation,or "no solution!", if there is no Sol Ution for the equation between 0 and 100.Sample Input

2100-4

Sample Output

1.6152No solution!

The code is as follows:

#include <iostream> #include <cmath> #include <iomanip>using namespace std;double y;double Equ (double x) {    return 8*pow (x,4) +7*pow (x,3) +2*pow (x,2) +3*x+6;             Determine if the equation is equal}double search ()                                               //Two min find {    double high=100,low=0;    Double temp;    while (high-low>1e-6)    {        temp= (Low+high)/2;        if (Equ (temp) <y)            low=temp+1e-7;        else            high=temp-1e-7;    }    Return (High+low)/2.0;} int main () {    int T;    cin>>t;    while (t--)    {        cin>>y;        if (Equ (0) <=y&&equ () >=y)                    //If the condition is met, a value of 0 to 100 must exist in the middle to make Equ (x) =y            cout<<setiosflags (ios::fixed) <<setprecision (4) <<search () <<endl;        else            cout<< "No solution!" <<endl;    }    return 0;}

Problem Solving Ideas:

The main idea is to find the x that satisfies 8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 = = In between 0 and 100.

Simple binary search, with emphasis on computational accuracy and output format.



Yt14-hdu-found the right x in 0-100.